Finding Kinetic Energy of Person Spinning on Chair w/ Weights

dvdqnoc

How do you find the Kinetic Energy of a person spinning on a chair with 2 equal weights on each arm?

A specific problem asks what the change in KE is if he moves his originally extended arms inward.

I tried doing (1/2)(I)(wf)^2 - (1/2)(I)(wi)^2 where wi is initial angular velocity and wf is final angular velocity, but its wrong.

Any help?

learningphysics

What numbers/variables do they give?

dvdqnoc

Ahh, sorry, I suppose I should have included this information in the main thread. My bad.

Well I is the same for initial and final, which is 8. Therefore,

Initial I = 8
Final I = 8
Initial w = .7
Final w = 1.15

So I did:

(1/2)(I)(wf)^2 - (1/2)(I)(wi)^2
= (1/2)(8)(1.15)^2 - (1/2)(8)(.7)^2
= 5.31118 - 1.96
= 3.35118116 j

...but apparently it's wrong.

learningphysics

I can't be the same initial and final... if he brings his arms in the moment of inertia changes. post the question exacty as it is stated...

dvdqnoc

Alright. But just fyi, I know it cant be the same I because the r changes, but the problem says to just assume its the same. I used the same I for initial and final to get the wf, and it was correct, so i know I is same initial and final. Anyways, heres the problem:


A student sits on a rotating stool holding two
3 kg objects. When his arms are extended
horizontally, the objects are 1 m from the axis
of rotation, and he rotates with angular speed
of 0.7 rad/sec. The moment of inertia of the
student plus the stool is 8 kgm^2 and is assumed
to be constant. The student then pulls
the objects horizontally to a radius 0.29 m
from the rotation axis.

a) Calculate the final angular speed of the
student. Answer in units of rad/s.
(The answer to this part is 1.15232 rad/s)

b) Calculate the change in kinetic energy of the
system. Answer in units of J.